The
probability of rolling a 7 with a pair of ordinary dice is 1/6, because of
the 36 possible (and presumably equally likely) combinations, six of them
yield the sum 7. Now, suppose the dice are rolled out of our sight, and some
honest person who can see the results tells us that at least one of the die
came up 6. This restricts the total number of possible combinations to the
following eleven pairs

(1,6) (2,6) (3,6) (4,6) (5,6) (6,6)
(6,1) (6,2) (6,3) (6,4) (6,5)

of
which exactly two yield a sum of 7, so we would now assess the probability of
a 7 as 2/11, just slightly better than 1/6. Of course, the same analysis
would give a probability of 2/11 if our honest friend had reported "at
least one 5" instead of "at least one 6". In fact, we would
arrive at the same probability if he reported "at least one n" for any
value of n. This sometimes strikes people as paradoxical, because obviously
we always have "at least one n" for some value of n, so why
not just assume this without waiting for our friend to tell us? It might seem
as if this magically improves our apriori odds of rolling a seven from
1/6 to 2/11.

This
is the kind of conundrum that often puzzles beginning students of
probability. It's actually a 6x6 version of the classical 2x2 problem that's
usually expressed in terms of boy/girl children. If we regard a man's two
children as a pair of 2-sided dice, each of which is equally likely to come
up as either a "1" (boy) or a "2" (girl), then what is
the probability of "total = 3"? The obvious answer is 1/2 (just as
the probability of rolling a 7 with two 6-sided dice is 1/6).

Now
we alter the problem by having the father provide us with one of two specific
pieces of (factual) information. He either tells us "At least one is a
boy" or he tells us "At least one is a girl". Given one of
these pieces of information, what is the probability of "total =
3"?

Strictly
speaking, the problem is underspecified in this form, because we don't know
how the father decides which piece of information to give us if he has a
choice (i.e., if he has one child of each gender). For example, suppose the
father tells us "At least one girl" whenever possible, and only
tells us "At least one boy" if he has no choice. Then obviously if
he tells us "At least one boy" we know the probability of
"total = 3" is zero. The key point here is the difference between

Condition
1: There is at least one boy.

Condition
2: There is at least one girl.

and

Condition
3: We are told that there is at least one boy.

Condition
4: We are told that there is at least one girl.

When
assessing the conditional probability of "total=3" we are not given
C1 or C2, we are given C3 or C4. C1 and C2 are each true exactly 3/4 of the
time, and half the time they are both true simultaneously. In contrast, C3
and C4 are mutually exclusive and complementary conditions, but we don't
really know what fraction of the time either of them is true. Therefore, the
problem is indeterminate.

However,
by the "unwritten rule of probability puzzles" (the UROPP), let's
assume that all specified choices are to be made randomly from the available
options. Thus we assume that if the man has one child of each gender he is
equally likely to report "At least one boy" as to report "At
least one girl". It follows that each of these reports has a probability
of 1/2. For clarity, let's denote C1 as [boys≥1], and C3 as [told:boys≥1].
On this basis we can use Bayes' formula to compute the conditional
probability as follows

The
denominator is known to be 1/2, and the numerator is 1/4 because the total
equals 3 exactly half the time, and (by assumption) the father reports
"At least one boy" exactly half of those times. Thus the
conditional probability of "total=3" given that the father tells us
"At least one boy" is exactly 1/2.

Similarly
in the 6x6 case we know that Pr{total=7} = 1/6, and by the UROPP we have

for
any n from 1 to 6, so Bayes formula gives the conditional probability

for
any n from 1 to 6. This in no way conflicts with the fact that

because
the conditions are different. To express this distinction verbally for the
2x2 case, suppose there are 100 fathers in an auditorium, and each is the
father of two children. Each father is instructed to tell us (truthfully) if
at least one of his children is a boy. This will apply to about 75 of the
fathers. Now, of those 75 Dads, 2/3 (i.e., 50) have a daughter, and 1/3
(i.e., 25) have two sons. Thus, if we want to guess the gender of their
"other" child, the chances are 2/3 that it is a girl. (Of course,
for the remaining 25 fathers - those who did not report at least one son - we
know immediately they have two daughters.)

However,
suppose instead that all 100 fathers were instructed to tell us either (a)
"At least one of my children is a boy" or (b) "At least one of
my children is a girl". Based on what each father tells us, we try to
guess the gender of his "other" child. Strictly speaking this
problem is indeterminate, but if it's also stipulated that fathers with both
a son and a daughter should flip a coin to decide what to tell us, then the
probability that the "other" child is of the opposite gender is
exactly 1/2. The breakdown is

25 have two
sons, and they report at least one son

25 have a son
and daughter, and report at least one son

25 have a son
and daughter, and report at least one daughter

25 have two
daughters, and they report at least one daughter

Thus,
regardless of what a particular father reports, we have only a 50% chance of
correctly guessing the gender of his "other" child.